Log and Antilog Amplifier MCQ [Free PDF] - Objective Question Answer for Log and Antilog Amplifier Quiz

11. Find the output voltage for the squarer circuit given below, choose input frequency as 10kHz and V_ref =10v

A. V_o = 5.0-(5.0×cos4π×10⁴t)
B. V_o = 2.75-(2.75×cos4π×10⁴t)
C. V_o = 1.25-(1.25×cos4π×10⁴t)
D. None of the mentioned

Answer: C

Output voltage of frequency V_o =V_i² / V_ref

=> V_i = 5sinωt = 5sin2π×10⁴t

V_o = [5×(sin2π×10⁴t)² ]/10

= 2.5×[1/2-(1/2cos2π ×2×10⁴t)] = 1.25-

(1.25×cos4π×10⁴t).

12. Calculate the phase difference between two input signals applied to a multiplier, if the input signals are V_x= 2sinωt and V_y= 4sin(ωt+θ). (Take V_ref= 12v).

A. θ = 1.019
B. θ = 30.626
C. θ = 13.87
D. θ = 45.667

Answer: A

V_o= [V_mx×V_my /(2×V_ref)] ×cosθ

=> (V_o×2×V_ref)/ (V_mx × V_my) = cosθ

=> cosθ = (10×2×12)/(2×4) = 30.

=> θ = cos^-130 =1.019.

13. Express the output voltage equation of the divider circuit

A. V_o= -(V_ref/2)×(V_z/V_x)
B. V_o= -(2×V_ref)×(V_z/V_x)
C. V_o= -(V_ref)×(V_z/V_x)
D. V_o= -V_ref²×(V_z/V_x)

Answer: C

The output voltage of the divider,

V_o= -V_ref×(V_z/V_x).

Where V_z –> dividend and V_x –> divisor.

14. Find the input current for the circuit given below.

A. I_Z = 0.5372mA
B. I_Z = 1.581mA
C. I_Z = 2.436mA
D. I_Z =9.347mA

Answer: B

Input current, I_Z = -(V_x×V_o)/(V_ref×R)

= −(4.79v×16.5v)/(10×5kΩ)

= 1.581mA.

15. Find the condition at which the output will not saturate?

A. V_x > 10v ; V_y > 10v
B. V_x < 10v ; V_y > 10v
C. V_x < 10v ; V_y < 10v
D. V_x > 10v ; V_y < 10v

Answer: C

In an analog multiplier, V_ref is internally set to 10v. As long as V_x < V_ref and V_y < V_ref, the output of the multiplier will not saturate.

16. Determine the relationship between log-antilog method.

A. lnV_x×lnV_y = ln(V_x+V_y)
B. lnV_x / lnV_y = ln(V_x-V_y)
C. lnV_x -lnV_y = ln(V_x/V_y)
D. lnV_x+ lnV_y = ln(V_x×V_y)

Answer: D

The log-analog method relies on the mathematical relationship that is the sum of the logarithm of the product of those numbers.

=> lnV_x+lnV_y = ln(V_x×V_y).

17. Which circuit allows doubling the frequency?

A. Frequency doubler
B. Square doubler
C. Double multiplier
D. All of the mentioned

Answer: A

The multiplication of two sine waves of the same frequency with different amplitude and phases allows for doubling a frequency.

18. Compute the output of the frequency doubler. If the inputs V_x = Vsinωt and V_y = Vision(ωt+θ) are applied to a four-quadrant multiplier?

A. V_o= { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/ V_ref
B. V_o= { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/ 2
C. V_o= { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×V_ref)
D. V_o= – { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×V_ref)

Answer: C

The output of frequency doubler

V_o= (V_x×V_y)/V_ref = [(Vsinωt +Vsin(ωt+θ)]/ V_ref

= {V_x×V_y×[(sin²ωt×cosθ)+(sinθ×sinωt×cosωt)]}/ V_ref

=> V_o= { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×V_ref).

19. How to remove the dc term produced along with the output in the frequency doubler?

A. Use a capacitor between load and output terminal
B. Use a resistor between load and output terminal
C. Use an Inductor between load and output terminal
D. Use a potentiometer between load and output terminal

Answer: A

The output of the frequency doubler contains a dc term and wave of double frequency. The dc term can be easily removed by using a 1µF coupling capacitor between the load and output terminal of the doubler circuit.

20. Find the voltage range at which the multiplier can be used as a squarer circuit?

A. 0 – V_in
B. V_ref – V_in
C. 0 – V_ref
D. All of the mentioned

Answer: C

The basic multiplier can be used to square any positive or negative number provided the number can be represented by a voltage between 0 to V_ref.

6. Calculate the output voltage of a squarer circuit, if its input voltage is 3.5v. Assume V_ref = 9.67v.

A. 2.86v
B. 1.27v
C. 10v
D. 4.3v

Answer: B

The output voltage of a squarer circuit

V_o = V_i² / V_ref

= (3.5)² / 9.67v = 1.27v.

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